Calc + applications in physics

The problem gives a cable weighing 2lb/fot, that is used to lift 800 lb of coal up a mineshaft 500ft deep. find the work

I looked up the answer, which 650,000 ft-lbs. The weight is the force, so there is no need for the acceleration due to gravity. I tried something that I thought was right, but it didn't yield the correct answer. Am I not integrating correctly or is my set-up wrong?

[tex] \int_{0 ft}^{500 ft} (800lbs + 2x) x\ dx [/tex]

I distributed the x and integrated, ending up with [tex] ({800/2})x^2 + ({2/3})x^3 [/tex] from 0 to 500.

Where did you get the 500 from... because won't that change when you pull the cable upwards?

Click to expand...

Remember for a constant force, the work is W=Fd. You can split the work up into the work for the coal (constant weight=constant force) + work for the cable. The force on the cable is the only thing that changes.